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Standard VIII

Chemistry

Isotopes and Isotones

Nuclidic mass...

Question

Nuclidic masses of $_{7}^{14} N$ and $_{7}^{15} N$ are mixed to give average atomic mass of 14.1. The ratio of $_{7}^{14} N$ and $_{7}^{15} N$ mixed is :

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Solution

Let ratio be $x : 1$ . Thus by law for average mass,
$x \times 14 + 15 = 14.1 (x + 1)$
$∴ x = 9$ Hence ratio of $_{7}^{14} N :_{7}^{15} N$ is $9 : 1$ or $9$

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Similar questions

What name is given to the pair of atoms such as $_{7}^{14} N a n d_{7}^{15} N$ ?

Q. The nucleodic masses of

^{14} N

and

^{15} N

are mixed to give atomic mass of 14.1. What is the ratio of the two isotopes as a whole number.

Q. The electrostatic energy of

Z

protons uniformly distributed throughout a spherical nucleus of radius

R

is given by

E = \frac{3}{5} \frac{Z (Z - 1) e^{2}}{4 π ε_{0} R}

The measured masses of the neutron,

_{1}^{1} H,_{7}^{15} N

and

_{8}^{15} O

are

1.008665 u, 1.007825 u, 15.000109 u

and

15.003065 u

respectively. Given that the radii of both the

_{7}^{15} N

and

_{8}^{15} O

nuclei are same,

1 u = 931.5 M e V / c^{2}

(

c

is the speed of light) and

e^{2} / (4 π ε_{0}) = 1.44 M e V f m

. Assuming that the difference between the binding energies of

_{7}^{15} N

and

_{8}^{15} O

is purely due to the electrostatic energy, the radius of either of the nuclei is

(1 f m = 10^{- 15} m)

$_{7}^{14} N$ weighs 14.003074 g and weighs 15.000 g. If average mass of Nitrogen is 14.0067, calculate the relative abundance of both isotopes ( $^{14} N$ and $^{15} N$ respectively).

Q. Obtain the binding energy (in MeV) of a nitrogen nucleus

(_{7}^{14} N)

[Given: m

(_{7}^{14} N)

=14.00307 u.]

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Nuclidic masses of 147N and 157N are mixed to give average atomic mass of 14.1. The ratio of 147N and 157N mixed is :

Let ratio be x:1. Thus by law for average mass,x×14+15=14.1(x+1)∴x=9 Hence ratio of 147N:157N is 9:1 or 9

Nuclidic masses of $_{7}^{14} N$ and $_{7}^{15} N$ are mixed to give average atomic mass of 14.1. The ratio of $_{7}^{14} N$ and $_{7}^{15} N$ mixed is :

Let ratio be $x : 1$ . Thus by law for average mass,
$x \times 14 + 15 = 14.1 (x + 1)$
$∴ x = 9$ Hence ratio of $_{7}^{14} N :_{7}^{15} N$ is $9 : 1$ or $9$